Question

Father of a family wishes to divide his square field bounded by x = 0, x = 4, y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them

Answer 1

Given curve y² = 4x and x² = 4y

Draw these two curves

Also draw the square bounded by the lines

x = 0, x = 4, y = 4 and y = 0

To prove Area A₁ = Area A₂ = Area A₃

Now the point of intersection of the curves y² = 4x and x² = 4y is given by

`(y^2/4)^2` = 4y

y⁴ = 64y

⇒ y(y³ – 64) = 0

y = 0, y = 4

when y = 0

⇒ x = 0

y = 4

⇒ x = 4

Point of intersection are O(0, 0) and B(4, 4)

Now, the area of the region bounded by the curves y² = 4x and x² = 4y is

A₂ = `int_0^4 (sqrt(4x) - x^2/4)  "d"x`

= `int_0^4 (2sqrt(x) - x^2/4)  "d"x`

= `[2  x^(3/2)/(3/2) - x^3/12]_0^4`

= `[4/3 (4)^(3/2) - 4^3/12 - 0]`

= `32/3 - 16/3`

= `16/3`   .......(1)

Now the area of the region bounded by the curves x² = 4y, x = 4 and x axis is

A₃ = `int_0^4 y  "d"x`

= `int_0^4(x^2/4)  "d"x`

= `[x^3/12]_0^4`

= `4^3/12 - 0`

= `16/3` sq.units   ......(2)

Similarly the area of the region bounded by the curve y² = 4x, y-axis and y = 4 is

A₁ = `int_0^4 x  "d"y`

= `int_0^4 (y^2/4)  "d"y`

= `[y^3/12]_0^4`

= `1/12 xx 4^3`

= `16/3`  ......(3)

Hence we see that

A₁ = A₂ = A₃ = `16/3` sq.units